Kinetics of heterogeneous catalytic reactions

The steps of a solid-catalysed reaction, rate bases, Langmuir adsorption, external and internal transport resistances and the packed-bed design equation, with coverage and resistance-in-series examples.

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Why it matters

Most large-tonnage chemicals — ammonia, sulphuric acid, methanol, cracked fuels — are made over solid catalysts. The observed rate in a catalytic reactor is the outcome of several steps in series: transport to the pellet, diffusion into pores, adsorption, surface reaction and desorption. Knowing which step controls tells you whether to change the catalyst, the pellet size, the flow rate or the temperature.

Key ideas

The seven steps. For a fluid reactant A reacting on a porous catalyst:

  1. External diffusion of A from the bulk fluid through the film to the pellet surface.
  2. Internal (pore) diffusion of A to the active sites.
  3. Adsorption of A on a site.
  4. Surface reaction of adsorbed species.
  5. Desorption of products.
  6. Internal diffusion of products to the pellet surface.
  7. External diffusion of products into the bulk. At steady state all steps proceed at the same rate; the slowest (highest resistance) is called rate-controlling and dominates the overall rate.

Rate bases. Catalytic rates are usually written per unit mass of catalyst, −r'A (mol/kg cat·s). Per unit bed volume: −rA = ρB·(−r'A), with ρB the bulk density of the bed (kg/m³). Per active site: turnover frequency (TOF), molecules converted per site per second.

Adsorption. Physisorption is weak (van der Waals, heat of adsorption roughly 10–40 kJ/mol), non-specific and multilayer. Chemisorption forms chemical bonds (often 40–400 kJ/mol), is specific, monolayer and activated — it is what catalysis needs. The Langmuir isotherm assumes a uniform surface, monolayer coverage, no interaction between adsorbed molecules and a fixed number of sites: θA = KA·pA/(1 + KA·pA). At low pressure coverage is proportional to pressure; at high pressure the surface saturates.

Surface-reaction-controlled rate laws. If adsorption and desorption are fast (at equilibrium) and the surface reaction is slow, the rate is proportional to the coverage of the reacting species. For A → products on single sites: −r'A = k·KA·pA/(1 + KA·pA + KR·pR). This gives apparent first order at low pA and zero order at high pA (details in the Langmuir–Hinshelwood topic).

Transport resistances.

  • External mass transfer: −r'A·ρB = kc·a·(CAb − CAs), where kc (m/s) comes from correlations (Sherwood number vs Reynolds and Schmidt) and a is the external surface per bed volume (m²/m³). It is reduced by higher velocity and smaller particles. When it controls, the rate barely depends on temperature (apparent E small) and depends on flow rate.
  • Internal diffusion: accounted for by the effectiveness factor η = actual rate/rate at surface conditions, which falls with pellet size (next-but-one topic). Strong pore diffusion halves the apparent activation energy.
  • Resistances in series (first order): 1/k_overall = 1/(kc·a) + 1/(η·k), all on a bed-volume basis. Diagnosis in the laboratory: change flow at constant space time (external effects), change particle size (internal effects), measure apparent activation energy.

Packed-bed design equation. In terms of catalyst mass W: W/FA0 = ∫dXA/(−r'A), the catalytic analogue of the PFR equation.

Formulas

−rA = ρB·(−r'A)

  • −r'A: rate per catalyst mass (mol/kg·s); ρB: bed bulk density (kg/m³).

θA = KA·pA/(1 + KA·pA) (Langmuir, single species)

  • KA: adsorption equilibrium constant (kPa⁻¹); pA: partial pressure (kPa); θA: fractional coverage.

θA = KA·pA/(1 + Σ Ki·pi) (competitive adsorption)

TOF = (molecules converted per second)/(number of active sites) (s⁻¹)

Rate = kc·a·(CAb − CAs) (external film transfer, per bed volume)

  • kc: mass-transfer coefficient (m/s); a: external area per bed volume (m²/m³); CAb, CAs: bulk and surface concentrations (mol/m³).

1/k_o = 1/(kc·a) + 1/(η·k) (first order, overall rate −rA = k_o·CAb)

  • k: intrinsic rate constant per bed volume (s⁻¹); η: effectiveness factor.

W = FA0·∫(0→XA) dXA/(−r'A) (packed-bed reactor)

  • W: catalyst mass (kg); FA0 in mol/s.

Worked examples

Example 1 (standard, Langmuir coverage). A gas adsorbs with KA = 0.02 kPa⁻¹. Find θA at pA = 100 kPa, and the pressure needed for θA = 0.9.

  1. θA = 0.02 × 100/(1 + 0.02 × 100) = 2/3 = 0.667.
  2. For θA = 0.9: KA·pA = θA/(1 − θA) = 9 → pA = 9/0.02 = 450 kPa.

θA = 0.667 at 100 kPa; 450 kPa needed for 90 % coverage.

Example 2 (GATE level, resistances in series). A first-order gas reaction in a packed bed has intrinsic k = 0.5 s⁻¹ (bed-volume basis), effectiveness factor η = 0.8, and external transfer kc·a = 0.25 s⁻¹. Find the overall rate constant, the share of resistance in the film, and the bed volume for 90 % conversion at v0 = 0.01 m³/s (plug flow, constant density).

  1. Film resistance: 1/(kc·a) = 1/0.25 = 4 s. Reaction-plus-pore resistance: 1/(η·k) = 1/0.4 = 2.5 s.
  2. k_o = 1/(4 + 2.5) = 0.1538 s⁻¹.
  3. Film share: 4/6.5 = 61.5 % — external mass transfer controls.
  4. Plug flow, first order: V = (v0/k_o)·ln[1/(1 − XA)] = (0.01/0.1538) × 2.303 = 0.150 m³.

k_o = 0.154 s⁻¹; film resistance 61.5 %; bed volume ≈ 0.150 m³. Raising the gas velocity (higher kc) would help far more than a more active catalyst.

Example 3 (TOF). A catalyst with 0.01 mol of active sites converts 0.5 mol/h: TOF = 0.5/0.01 = 50 h⁻¹ = 0.0139 s⁻¹.

Common mistakes

  • Mixing rate bases (per kg catalyst vs per m³ bed) without the bulk density.
  • Assuming adsorption is always fast; the controlling step must be tested against data.
  • Treating a rate that is insensitive to temperature as "low activation energy chemistry" when it is really film-diffusion control.
  • Adding rate constants instead of resistances for steps in series.
  • Using pressure in atm in one constant and kPa in another.

For GATE CH

Questions cover the sequence of steps and identifying the controlling step from experimental clues, Langmuir coverage calculations, overall rate constants from resistances in series, rate-basis conversions, and packed-bed catalyst mass from the design equation. Practise the resistance-in-series algebra and the diagnostic tests (flow rate, particle size, temperature).

Quick check

  1. Which adsorption type is responsible for catalysis?
  2. If the observed rate rises when gas velocity increases at constant W/FA0, which step is limiting?
  3. With KA·pA ≫ 1, what is the apparent order of a single-site surface reaction in A?
  4. Convert −r'A = 0.002 mol/kg·s to a bed-volume rate for ρB = 800 kg/m³.

Answers: 1. chemisorption; 2. external mass transfer; 3. zero order; 4. 1.6 mol/m³·s.

Interview questions

All Chemical Reaction Engineering interview questions

Try answering each one aloud before you open it.

  1. 1.What is a heterogeneous catalytic reaction?Concept

    A heterogeneous catalytic reaction is a chemical reaction where the catalyst is in a different phase than the reactants. Typically, the catalyst is a solid, while the reactants are gases or liquids. This type of reaction occurs on the surface of the catalyst, where the reactants adsorb, react, and then desorb as products.

  2. 2.Explain the role of adsorption in heterogeneous catalytic reactions.Concept

    Adsorption is a crucial step in heterogeneous catalytic reactions. It involves the binding of reactant molecules to the surface of the catalyst. This process increases the concentration of reactants at the catalyst surface, facilitating the reaction. Adsorption can be either physisorption, which is weak and reversible, or chemisorption, which is stronger and often involves the formation of chemical bonds.

  3. 3.Why is surface area important in heterogeneous catalysis?Application

    Surface area is important in heterogeneous catalysis because it determines the number of active sites available for the reaction. A larger surface area provides more sites for adsorption, leading to higher reaction rates. Catalysts are often designed to have high surface areas to maximize their efficiency.

  4. 4.What happens if the catalyst in a heterogeneous reaction is poisoned?Application

    If a catalyst is poisoned, its active sites are blocked by impurities or unwanted substances, reducing its effectiveness. This can lead to a decrease in reaction rate or even complete deactivation of the catalyst. Catalyst poisoning is a significant issue in industrial processes and requires careful management to maintain efficiency.

  5. 5.Explain the difference between physisorption and chemisorption in the context of catalysis.Concept

    Physisorption involves weak van der Waals forces and is usually reversible, with no significant change in the electronic structure of the adsorbate. Chemisorption, on the other hand, involves the formation of chemical bonds between the adsorbate and the surface, often leading to a change in the electronic structure. Chemisorption is usually stronger and more specific than physisorption, making it more relevant in catalytic processes.

  6. 6.Why are transition metals commonly used as catalysts in heterogeneous reactions?Application

    Transition metals are commonly used as catalysts because they have partially filled d-orbitals that can form bonds with reactants. This allows them to facilitate the breaking and forming of chemical bonds during the reaction. Additionally, transition metals can exhibit multiple oxidation states, providing flexibility in catalyzing various reactions.

  7. 7.What is the effect of temperature on the rate of a heterogeneous catalytic reaction?Application

    The rate of a heterogeneous catalytic reaction generally increases with temperature due to higher kinetic energy of the reactants, leading to more frequent and energetic collisions. However, excessively high temperatures can lead to catalyst sintering, where the catalyst particles agglomerate, reducing surface area and activity.

  8. 8.Calculate the turnover frequency (TOF) if a catalyst converts 0.5 mol of reactant per hour and has 0.01 mol of active sites.Numerical

    TOF is calculated as the number of moles of reactant converted per mole of active sites per unit time. Here, TOF = (0.5 mol/h) / (0.01 mol) = 50 h⁻¹.

  9. 9.A catalytic reaction has an activation energy of 50 kJ/mol. If the temperature is increased from 300 K to 350 K, by what factor does the rate constant change, assuming the Arrhenius equation applies?Numerical

    k2/k1 = exp[(E/R)·(1/T1 − 1/T2)] = exp[(50 000/8.314) × (1/300 − 1/350)] = exp(6014 × 4.762 × 10⁻⁴) = exp(2.864) ≈ 17.5. So the intrinsic rate constant rises about 17.5 times. In a catalytic reactor the observed rate may rise much less if pore diffusion (apparent E roughly halved) or film mass transfer (apparent E very small) becomes controlling at the higher temperature.

  10. 10.What is the significance of the Langmuir-Hinshelwood mechanism in heterogeneous catalysis?Concept

    The Langmuir-Hinshelwood mechanism describes a model where both reactants adsorb onto the catalyst surface before reacting. It is significant because it provides a framework for understanding how surface reactions occur and helps in deriving rate equations for reactions involving adsorbed species. This mechanism is widely used to explain the kinetics of many catalytic processes.

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